Is the Entropy of Black Holes Measurable for Non-Exotic Black Holes?

In summary, my physics professor told us that string theory correctly predicts the entropy of a black hole. that leaves me wondering...how do u even measure what it's entropy is to even confirm a theoreticla calculation?
  • #1
RasslinGod
117
0
My physics professor told us that string theory correctly predicts the entropy ofa black hoole. that leaves me wondering...how do u even measure what it's entropy is to even confirm a theoreticla calculation?

is S= k ln W even used at all?
 
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  • #3
RasslinGod said:
My physics professor told us that string theory correctly predicts the entropy ofa black hoole. that leaves me wondering...how do u even measure what it's entropy is to even confirm a theoreticla calculation?

is S= k ln W even used at all?

As far I know, this relation has been used only for "exotic" black holes, i.e., black holes that have one or more of supersymmetry, Yang-Mills charge, and extra macroscopic spatial dimensions.

Also as far as I know, astrophysical black holes have none of these properties.
 
  • #4
Recently people in strings have been doing non-supersymmetric as well.. of course these are not even close to astorphysical black holes...
 
  • #5
Hawking and Unruh effect

There is the Hawking-Unruh effect, which tend to merge in a way for an accelerated frame near a black hole. To examine this it is best to transform to another set of coordinates to look at this. The Kruskal-Szekeres coordinates are better. These are given by

[tex]
u~=~\sqrt(r/2M - 1)e^{r/4M}cosh(t/2M),
[/tex]
[tex]
v~=~\sqrt(r/2M - 1)e^{r/4M}sinh(t/2M).
[/tex]

It is clear that

[tex]
(r/2M~-~1)e^{r/2M}~=~u^2~-~v^2
[/tex]
and

[tex]
tanh(t/4M)~=~v/u.
[/tex]

This has the advantage that there is no funny business at [itex]r~=~2M[/itex]. These coordinates provide a chart which covers the whole space. The [itex]u^2~-~v^2[/itex] equation shows that the inner and outer regions of the black hole are given by a branch cut connecting two sheets.

Consider [itex](r/2M~-~1)e^{r/2M}~=~x^2[/itex], and so [itex]x^2~=~u^2~-~v^2[/itex]. The u and v satisfy hyperbolic equations for a constant r and so

[tex]
u~=~sinh(gs),~ v~=~cosh(gs).
[/tex]

It is clear that for [itex]udu~-~vdv~=~e^{r/2M}/4M dr[/itex] that [itex]g~=~1/(r/2M~-~1)e^{t/2M}/4M[/itex] , and so the accleration diverges as [itex]r~\itex~2m[/itex] in order to hold a particle fixed near the event horizon.

The hyperbolic equations above are then identical in form to the ones which obtain for the Unruh effect. The spacetime trajectory in the u and v coordinates is then hyperbolic. Thus for an observer sitting on a frame fixed next to the event horizon of the black hole there would be a huge thermal bath of particles which would become very hot [itex]T~\rightarrow~\infty[/itex] as [itex]r~\rightarrow~2M[/itex]. The four velocities [itex]U_u[/itex] and [itex]U_v[/itex] will then satisfy

[tex]
U_u = 1/(4M(r/2M - 1))e^{t/2M}cosh(gs), U_v = 1/4M((r/2M - 1))e^{t/2M}sinh(gs),
[/tex]

with

[tex]
(U^u)^2~-~(U_v)^2~=~16M^2(1~-~r/2M)e^{-t/2M},
[/tex]

and so the temperature of the black hole is [itex]T~\sim~1/2\pi g[/itex] and

[tex]
T~=~\frac {e^{t/M}}{8\pi M}\sqrt{1~-~r/2M}).
[/tex]

Now for [itex]r~\rightarrow~\infty[/itex] and [itex]t~\rightarrow~\infty[/itex] this recovers the standard black hole temperature result of

[tex]
T~=~\frac{1}{8\pi M}.
[/tex]

Thus for an observer held close to the event horizon of a black the additional acceleration effectively heats up the black hole on that frame.

I think this might also indicate that for a distant observer there is a question as to what she would see of her compatriot held fixed near the black hole. This might be an interesting problem to examine.
 
  • #6
I made a couple of Tex errors on the above discussion which I corrected

L. C.
 
  • #7
George Jones said:
As far I know, this relation has been used only for "exotic" black holes, i.e., black holes that have one or more of supersymmetry, Yang-Mills charge, and extra macroscopic spatial dimensions.

Also as far as I know, astrophysical black holes have none of these properties.

These exotic black holes are BPS black holes, and pertain to issues of quantum gravity or dualities between D-branes and gauge charges. The astrophysical black hole is massive and these putative physics are in some infinitesimal region (or a small number of Planck lengths) from the event horizon. Hence for a stellar black hole these BPS things don't contribute much of anything.

Lawrence B. Crowell
 

What is entropy of a black hole?

Entropy is a measure of the disorder or randomness in a system. In the context of black holes, it is a measure of the number of microstates or ways in which the information of the matter inside the black hole can be arranged.

How is entropy related to black holes?

According to the laws of thermodynamics, the entropy of a closed system, such as a black hole, always increases or remains constant. This means that as matter falls into a black hole and its mass increases, the entropy also increases.

How is the entropy of a black hole calculated?

The entropy of a black hole can be calculated using the famous equation proposed by physicist Jacob Bekenstein. The equation states that the entropy is equal to one-quarter of the black hole's event horizon area divided by the Planck length squared.

Can the entropy of a black hole decrease?

No, according to the second law of thermodynamics, the entropy of a closed system can never decrease. Therefore, the entropy of a black hole can only increase or remain constant.

What does the entropy of a black hole tell us about its information?

The entropy of a black hole is directly related to the amount of information it contains. As the entropy increases, the amount of information also increases. It is believed that the information of matter that falls into a black hole is encoded on its event horizon, also known as the holographic principle.

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